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 April 7th, 2013, 06:37 PM #1 Member   Joined: Jan 2013 Posts: 54 Thanks: 0 Convergence and divergence 1. infinity E.........1/n n = 1 2. infinity E ........1/n^2 n = 1 The first series diverges and the second converges to pi^2/6. Somewhere between the two series, behaviour switches from divergence to convergence. Where is the dividing line? First, I don't get why the first series diverges and the second series converges. They are both approaching 0 but the second series is approaching it faster. Why is this so?
April 8th, 2013, 12:56 AM   #2
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Re: Convergence and divergence

Quote:
 Originally Posted by math221 Where is the dividing line?
The question is somewhat vague. Depending on what you want will give you different answers. For example, we could say

$\Re[z] > 1$

Is the dividing line.

Quote:
 Originally Posted by math221 I don't get why the first series diverges and the second series converges.
The first series is just not fast enough to converge. Furthermore, taking the set of primes instead of the integers diverges as O(log(log n)) forcing the more general sum to diverge (like log(n)).

Quote:
 Originally Posted by math221 They are both approaching 0 but the second series is approaching it faster.
The limit test is a necessary but not sufficient to test convergence.

 April 8th, 2013, 03:41 AM #3 Math Team   Joined: Nov 2010 From: Greece, Thessaloniki Posts: 1,989 Thanks: 133 Math Focus: pre pre pre pre pre pre pre pre pre pre pre pre calculus Re: Convergence and divergence [color=#000000]$\sum_{n=1}^{\infty}\frac{1}{n}$ is divergent because $\lim_{R\to\infty}\int_{1}^{R}\frac{1}{t}\;\mathbb{ d}t=\lim_{R\to \infty}\ln(R)=+\infty$. $\sum_{n=1}^{\infty}\frac{1}{n^2}$ is convergent because $\lim_{R\to\infty}\int_{1}^{R}\frac{1}{t^2}\;\mathb b{d}t=\lim_{R\to \infty}1-\frac{1}{R}=1<\infty=$.[/color]
 April 8th, 2013, 07:39 AM #4 Math Team   Joined: Sep 2007 Posts: 2,409 Thanks: 6 Re: Convergence and divergence By the ratio test, $\sum_{n=1}^\infty \frac{1}{n^\alpha}$ converges as long as $\alpha> 1$ so that "1" is the "dividing line".

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